H-H H + H 430 kJ + Breaking bond always requires E Endothermic: - - PDF document

SLIDE 1

1

Topic 9.2 Heat of reaction, heat of formation, Table I

H-H H + H H + H H- H

430 kJ + + 430 kJ

Breaking bond always requires E

• Endothermic: heat = R
• E absorbed

Forming bonds always releases E

• Exothermic: heat = P
• E released

Heat of formation =

ΔH for 1 mole of comp’d formed from elements in standard conditions (25°C and 1 atm)

Heats of solution =

Types of Heats of Reaction (ΔH)

ΔH for the dissolution of a soluble comp’d

NaCl (s) H 2O (g) H2 (g) H2 (g)+

1 2 O 2 (g) H 2O (g)

NaCl (s) Na

+ (aq)

NaCl (s) Na

+ (aq) +Cl

• (aq)

ΔH for combustion rxn

Heats of formation Heats of solution for ionic compounds ΔH for neutralization rxn

Formation of liquid water

2H2(

g ) + O2(g) 2H2O(l )

2H2O (l ) 2H2

(g) + O2(

g )

ΔH = ΔH = -571.6 kJ

+ 571.6 kJ

571.6 kJ +

If rxn reversed, change sign for ΔH

Decomposition of liquid water

+571.6 kJ

1 2H2(g ) + 1 4 O2(g) 1 2 H2O(l)

l)H = 142.9 kJ

What ever you do to coefficients do to ΔH

Divide ΔH by 4

1) What is the heat of formation for 0.5 mole H2O(l)?

2H2( g ) + O2(g) 2H2O(l)

From table:

l)H = 571.6 kJ

SLIDE 2

2

2) What is the heat of formation for 0.2 mole H2O(g)?

l)H = 48.3 kJ

Divide ΔH by 10

2H2(

g ) + O2(g) 2H2O(

g )

From table:

l)H = 483.6 kJ

3) What is the heat of reaction for the combustion of 8 g mole C6H12O6?

C6H12O6 + 6O2(g) 6CO2(g) + 6H2O(l)

From table:

H = 2804 kJ

8 g

H =

8 g 1 mole C6H

12O6

180 g

• =
• = 0.044 mole C6H

12O6

(0.044)( )(2804) ) = 123 kJ

4) One mole of which of the following compounds would react with oxygen to release the most energy?

C8H18(l) C6H12O6(s) C(s) H2(g)

2 C8H

18 + 25 O2 16 CO2 +18 H2O

2 C6H

12O6 + 6 O2 6 CO2 + 6 H2O

C + O2 CO2

2 2

2 H2 + O2 2 H2O O H = -10,943 kJ O H = - 2804 kJ H = - 393 kJ O H = - 483 kJ

5) In which reactions do the products have a lower energy content than the reactants? Exothermic a) Combustion of sugar b) Formation of NO2 c) Decomposition of liquid water d) Dissolution of NaOH 2C6H

12O6 + 6O2 6CO2 + 6H2O H = - 2804 kJ

2H2O(l) 2H2(g) + O2(g) N2 + 2O2 2NO2 H = +66.4 kJ NaOH(s)2

H2O

• Na+

(aq) + OH (aq) H = -44.5 kJ

H = +571.6 kJ

Using ΔH formation to calculate ΔH reaction:

6) Calculate the ΔH reaction for the reaction of nitrogen monoxide with oxygen to yield nitrogen dioxide. balanced equation:

2NO(g) +

The ΔHf for any element in standard state = +66.4/2 kJ +182.6/2 kJ

2NO2(g) 2 2NO Hrxn = = +66.4

Find ΔHf values from table I:

+ O2(g) Hfprod.

• prod. Hfreact.

0 kJ

182.6 = 116.2 kJ

for 2 mol NO2

6 = 58.1 kJ

for 1 mol NO2

7) Calculate the ΔH reaction for the combustion of ethyne balanced equation:

• 393.5 kJ

227.4 kJ 0 kJ

• 571.6 kJ

Find ΔHf values for 1 mole from table I:

C2H2(g) + O2(g) CO2(g) + H2O(l) 2.5

Hrxn =

Hrxn =

2

1/2( ) 2( )

Hfprod Hfreact

2(393.5) ) +1/ 2(571.6) ) 1(227.4) ) = 1300.2 kJ

SLIDE 3

3

Calculate the ΔH reaction for the 8) combustion of C2H4(g) 9) vaporization of liquid water